Bessler's (4th) Kassel wheel Archimedes screw pump calculations

[MrTim]

The only assumptions for the hammer-mill is how high the weights were lifted and the density weight of the wood.

The Merseburg drawing shows the stampers with the notches in them (and also the 'teeth' on the shaft that fit into them.)
The Weissenstein drawing also shows stampers, with the notches also visible.
Shouldn't be too hard to determine the distance they could move from the drawings...?

As far as density of wood, wouldn't the heaviest be used? (Mainly because those who would build and use such machinery back then, and saw the wheel in the demonstrations, would be less than impressed if a light weight wood were used, and would also feel that somebody was trying to put one over on them.)

[daanopperman]

Hi jim_mich ,

Just a simple question , would it matter if the pitch was different in 2 identical tubes , I seem to think that with a smaller pitch , the level in the scoops would drop , but you would have more , and in a larger pitch , the level in the scoops would just rise to capacity , that would be a level mark up to the shaft from the top of the last scoop .
I think you could calculate the amount of water in the tube by simply taking a level mark from the discharge to the wall of the tube , then down vertically till you hit the bottom of the tube , then horizontally backwards and carry on till you exit the tube .

[Unbalanced]

I belong to a historical drinking society or is that a drinking historical society, regardless, ECV as we are broadly known, document historical sites and errect monuments and often restore old things we find out in the bushes, mostly in the states of California and Nevada.

Here is a two stamp mill we restored some time back and I post it here to demonstrate the amount of power that was required to run one of these mills. I don't know what horsepower the steam engine was.

On a side note, any calculation as to the weight of a single of JB's stamps should include the weight of an iron shoe for the stamp and an iron piece that engaged the lifting cog.

http://m.youtube.com/watch?v=gbDeH9xCcm8

[daanopperman]

jim-mich ,

Maybe I am wrong , but the amount of scoops does not matter , in all cases , the screw is half full and half empty , as long as the pump suction is fully submerged .

[ovyyus]

Jim, if the Kassel wheel could lift a hundred weight directly from its 8 inch axle at a constant 20 RPM (or greater) then it seems obvious that a much larger braking force must be applied to the axle in order to completely stop it. How was the wheel stopped by its operator if it was as powerful as you believe? What is your estimate of the method and effort required to stop the wheel?

If the wheel developed around 1/4 horsepower then how was it controlled by the operator? Also, why would Leibniz say the wheel was very weak and of no practical value if it actually developed around 1/4 horsepower as you believe?

Dan, the difference is related to RPM and the volume of water lifted in a given time.

[jim_mich]

As far as density of wood, wouldn't the heaviest be used?

Yes.
Bessler gives us the dimensions of the stamper hammers of the Kassel wheel. In order to know their weight we must assume the type of wood. (I have data on wood densities.)

I feel European red oak would be a reasonable choice. Ebony would make the wheel output 66% greater. Hemlock (very unlikely) would make the wheel output 47% less.

daanopperman, the water screw must have an air-gap near its axis, else it runs the risk of siphoning. This limits the amount of water in each "scoop" to less than 50%. The percentage of water relative volume within the screw depends upon a number of things. It can run as low as 20% with a steep tilt of the screw, or as high as 40% or more with a very shallow angle. When the screw lead is short there ends up a greater percentage of water in the screw. But less water exits with each rotation. If the screw lead was doubled, say from 6 inches to 12 inches, then a greater amount of water is pumped. But the volume does not double because there is less water per scoop.

[daxwc]

Wagner:

One can also readily conclude that his internal motive principle or so-called superior force has no hundredweight force because his apprentice - a weakling who together with his leather waistcoat and blue apron does not weigh a hundredweight but is able to halt the wheel by grasping the axle twice - made it clear that the wheel cannot raise a hundredweight; for a body weighing less than a hundredweight suffices to bring the wheel to a complete standstill within 2 seconds, and as soon as the wheel stops moving, almost all internal impetus is gone.

[daxwc]

Jim:

daanopperman, the water screw must have an air-gap near its axis, else it runs the risk of siphoning.

Explain please? You can’t be worried about siphoning up, because that would be to your advantage. Your worried about water siphoning down out of the screw?

[Unbalanced]

The steam engine originally used to power the two stamp mill in the video I posted above is 10 HP. The boiler came off a wrecked locamotive. To turn a four stamp mill in a single revolution of a wheel would require at least this, I would assume but as a qualifier, I know dick about this stuff.

[jim_mich]

If the wheel developed around 1/4 horsepower then how was it controlled by the operator? Also, why would Leibniz say the wheel was very weak and of no practical value if it actually developed around 1/4 horsepower as you believe?

Where are you getting this 1/4 HP? My calculations show about halfway between 1/5 and 1/6 HP (about 0.1848 HP).

That 0.1848 HPt is a force of 112 pound at 8 inch diameter. At the outside of the 12 foot diameter wheel, the force would be about 6-1/4 lbs. Can you exert a push of 6-1/4 pounds? The momentum of the wheel would require much more stopping force than the 6-1/2 lbs.

If you assume a specific wheel mass and assume the weight distribution (near the rim or nearer the axle) and assume how long the force is applied (such as how many rotations it takes for the wheel to reach a given speed) then we can calculate how much force is needed at the rim (or at any other place on the wheel) to bring the wheel up to speed. This same force is needed to stop the wheel, plus whatever constant force the 6-1/4 pounds mentioned above is causing.

[daxwc]

Construction Principles



How efficient an Archimedes screw works is determined by its structure: the diameter (inside and out), the angle it’s placed at and how long it is, the angle of the blades, the number of blades and the rate at which the internal component is turned.

When water moves down a pipe, the rate of flow is faster in the central part of the pipe than toward the outer diameter. This rate of flow is a consideration when designing an Archimedes screw for use with an application. If the use of the screw is to force material from one source to another, then the distance between the interior diameter of the cylinder and the screw blades needs to be very small. If there is not the need for high pressure, then there can be a more open area between the blades and the interior wall of the cylinder.


Operation Principles



The speed at which the screw is turned has to allow for a smooth movement of the material being moved. If too fast, the water doesn’t have time to fill the open areas within the cylinder before the screw turns, reducing the efficiency of the device.

Also, just as it is more of a challenge to carry a full bucket of water than an empty one, the Archimedes screw becomes more difficult to turn when there is more water in it; the pressure needed to turn it increases.

Read more: http://www.ehow.com/about_6389901_do-ar ... z2jdHkdESY

[jim_mich]

I think Wagner got his facts screwed up a little. I think the only wheel that Wagner ever saw was the Merseburg wheel (I may be wrong)

How long did the handle stick out from the 6 inch diameter axle? Say it stuck out 12 inches, making the end of the handle 15 inch radius from wheel center. The load of the Merseburg wheel was about 70 to 80 pound acting at 3 inch radius. That same force acting at 15 inch radius would be 14 to 16 pound. Do you think Bessler's assistant might apply 70 to 80 pounds of counteracting force to stop the wheel momentum and counter the 14 to 16 lbs of driving force?

Of course he could.

[Unbalanced]

Even if we assume that the wooden hammers of Bessler's were considerably lighter than those in the mill shown in the video I posted above, it is extremely unlikely (in my estimation) that 1/4 horse power would be adequate power to lift twice as many hammers per revolution.

The 10 HP steam engine used in this video was the original purchased for the purpose.

This leads me to believe that either your calculations are far from the reality of the wheel's actual power output, or that it is not realistic to assume that the wheel could actually lift four hammers per revolution.

[ovyyus]

Where are you getting this 1/4 HP?

You estimated wheel power at around 160 Watts. That's about 1/4 HP.

[jim_mich]

Bill, when and where did I estimate wheel power at around 160 Watts? I might have, but I really don't think so.

I think if you do a little checking you will find my estimates have always been somewhat less than 160 Watts.

These are my numbers...

Kassel (4th) Wheel
Lifting 112 lbs:
HP = 0.1848145 Horse Power (almost 1/5 HP)
W = 137.8 Watts at 26 RPM.
W = 106 Watts at 20 RPM
Kassel (4th) Lifting Hammers:
HP = 0.164306 Horse Power (almost 1/6 HP)
W = 122.5 Watts at 26 RPM
W = 94.2 Watts at 20 RPM

Merseburg (3rd) Wheel
Lifting 80 lbs at 50 RPM:
HP = 0.1665996 Horse Power (about 1/6 HP)
W = 124.233 Watts
Lifting 70 lbs at 40 RPM:
HP = 0.11661972 Horse Power (between 1/8 and 1/9 HP)
W = 86.963 Watts
Lifting Hammers at 50 RPM:
HP = 0.149697 Horse Power (almost 1/7 HP)
W = 111.6 Watts
Lifting Hammers at 40 RPM:
HP = 0.1197576 Horse Power (almost 1/8 HP)
W = 89.3 Watts